I've got stuck on an induction problem, any help would be great

A(1)=√5 A(n+1)=√5*a(n)

The terms of the sequence are: √5, √(5√5), √(5(√5(√5))),....

Use induction to show that for natural number n, 0<A(n)<5

Printable View

- May 8th 2010, 06:57 AMiwish123Induction
I've got stuck on an induction problem, any help would be great

A(1)=√5 A(n+1)=√5*a(n)

The terms of the sequence are: √5, √(5√5), √(5(√5(√5))),....

Use induction to show that for natural number n, 0<A(n)<5 - May 8th 2010, 08:09 AMArchie Meade
Hi iwish123,

$\displaystyle A_1=\sqrt{5}$

$\displaystyle A_{n+1}=\sqrt{5A_n}$

Show by induction that $\displaystyle 0<A_n<5$

If $\displaystyle A_n\ <\ 5$

then $\displaystyle \sqrt{A_n}\ <\ \sqrt{5}$

hence $\displaystyle A_{n+1}=\sqrt{5A_n}=\sqrt{5}\sqrt{A_n}\ <\ \sqrt{5}\sqrt{5}\ <\ 5$

As $\displaystyle A_1=\sqrt{5}$

then

$\displaystyle A_2<5,\ A_3<5,\ A_4<5.......$

So, $\displaystyle A_1$ being < 5__causes__all subsequent terms of the sequence to be < 5 also.

A non-inductive proof is as follows...

$\displaystyle A_1=\sqrt{5}$

$\displaystyle A_2=\sqrt{5\sqrt{5}}=\sqrt{5}\sqrt{\sqrt{5}}=5^{\f rac{1}{2}}5^{\frac{1}{4}}$

$\displaystyle A_3=\sqrt{5\sqrt{5\sqrt{5}}}=\sqrt{5}\sqrt{\sqrt{5 }}\sqrt{\sqrt{\sqrt{5}}}=5^{\frac{1}{2}}5^{\frac{1 }{4}}5^{\frac{1}{8}}$

$\displaystyle A_n=5^{\frac{1}{2}}5^{\frac{1}{4}}5^{\frac{1}{8}}. .....$

$\displaystyle =5^{\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\fra c{1}{16}+....\right)}=5^{\frac{1}{2}\left(1+\frac{ 1}{2}+\frac{1}{4}+\frac{1}{8}+....\right)}$

The index is a geometric series, first term=1, common ratio = 0.5 (for the part in brackets)

$\displaystyle S_{\infty}=\frac{1}{1-\frac{1}{2}}=2$

so the index of 5 is 1.

Hence $\displaystyle A_n$ cannot reach 5 as a finite number of terms will not reach the sum to infinity as all the terms are positive. - May 8th 2010, 08:21 AMiwish123
Thanks, thats a great help

How would I go about proving that the sequence is convergent, through showing its an increasing sequence and bounded? I assume that I can say it is bounded as its been shown that the sequence is less than 5. - May 8th 2010, 08:31 AMArchie Meade